Finally we are ready to begin the study of the generalization of the Weihrauch hierarchy toκκ andRκ. In
this section we will focus on choice principles.
Choice principles have a central role in classical Weihrauch reducibility theory. In particular, as shown in [5], one can use them to characterize different Weihrauch degrees which are of main interest from the computable analysis point of view. In this section we will generalize some of these choice principles and we will start their classification within the Weihrauch hierarchy. For a complete introduction to classical choice principles we refer the reader to [5] and [4]. Before we begin the study of generalized choice principles, let us generalize the definition of Weihrauch reducibility and fix some conventions.
Definition 4.5.1 (Realizers). Let F :⊆ M1 ⇒ M0 be a multi-valued function over the represented spaces
(M1, δM1)and(M0, δM0). Thenf :⊆κ
κ→κκ is a realizer of F iff for everyx∈dom(F◦δ
M1)we have
δM0(f(x))∈F(δM1(x)).
Definition 4.5.2 (Generalized Weihrauch Reducibility). Let F :⊆M1 ⇒ M0 and G:⊆N1 ⇒ N0 be two
multi-valued functions between represented spaces. Then we will say that F is Weihrauch reducible toG, in symbols F ≤w G, iff there are two (partial) continuous functions H :⊆κκ →κκ and K :⊆κκ →κκ such
that for every realizerg:⊆κκ→κκ of Gthere is a realizer f :⊆κκ→κκ ofF such that
f =H◦ dID, g◦Ke,
where ID : κκ → κκ is the identity function. Moreover, if F and G are such that for every realizer g :⊆
κκ→κκ of Gthere is a realizer f :⊆κκ→κκ ofF such that
f =H◦g◦K,
then we will say that F is strongly Weihrauch reducibletoG, in symbolsF ≤s,wG.
As usual if F is (strongly) Weihrauch reducible toGandGis (strongly) Weihrauch reducible toF then we will say thatF is (strongly) Weihrauch equivalenttoGand we will writeF ≡wG(F ≡s,wG).
From now on we will considerRκ,R<κ andR >
κ as represented byδRκ,δR<κ andδR>κ, respectively. Moreover,
κwill be represented by
δκ(p) =p(0).
LetP−(κ) be the power set ofκminusκitself. We will represent the hyperspaceP−(κ) by
δP−(κ)(p) ={α| ∀β. p(β)6=α}.
Finally, the set Cl[0,1] of closed subsets of [0,1] will be represented by the following function:
δ[0,1](p) =A⇔A=Rκ\
[
{I0(w)|ι(w)/ p},
where I0:κ<κ→κis defined as follows:
I0(w) = (q1, q2) iffνQ−1
κ×Qκ(w) =dq1, q2e.
Namelypis a code forAiff it is a list of open intervals inRκwithκ-rational end points whose union is the
complement ofA.
Definition 4.5.3 (Choice Principles). We will consider the Weihrauch degrees of the following choice prin- ciples:
• Cκ
I interval choice: given a non empty closed interval in [0,1], the function CκI chooses an element in
the interval. Formally Cκ
I is defined as the following multi-valued function:
CκI : Cl[0,1]⇒Rκ dom(CκI) ={[a, b]|0≤a≤b≤1}.
• Cκ discrete choice: given a non empty subset of κ, the function Cκ chooses an element of the set.
Formally Cκ is defined as the following multi-valued function:
Cκ:P−(κ)→κ dom(Cκ) ={A⊂κ|A6=∅}.
• For every U ⊆κκ we consider the following multi-valued function C
U :κκ⇒κκ given by: CU(p)(α) = ( 1 (p)α∈U, 0 otherwise, where (p)α(β) =p(dα, βe).
In this section we will begin the study of the diagram in Figure 4.23. First of all we will show that Cκ and CκI are not continuous:
.. . CUα+1 CUα O O .. . CU2 O O IVTκ CU1 O O Cκ I ≡wBκI O O / / //Cκ≡wBκ j j \ ~ ~ ID ` ` >>
Figure 4.2: A part of the Weihrauch hierarchy.
Proposition 4.5.4. The following hold:
ID<wCκ andID<wCκI.
Proof. Note that the identity ID can be trivially reduced to any function, so it is enough to show that Cκ
and CκI do not reduce to ID.
We will prove Cκ6≤wID. By contradiction assume Cκ≤wID. Then there are two continuous functions
H and K such thatH◦ dID,ID◦Keis a realizer of Cκ. Note that, since ID, H andK are continuous we
have that
H◦ dID,ID◦Ke
is continuous. Therefore, there is a Wadge strategyθ:⊆κ<κ→κ<κ such that for every p∈κκ
H◦ dID,ID◦Ke(p) = [
α∈κ
θ(pα).
Now, letα∈κbe such that
δκ([θ(pα)])⊆X
whereX ⊂κ. Definep00= (pα)_p0 where p0 is any map in κκ such thatX ⊆ {β∈κ| ∃α < κ. p(α) =β}. By the monotonicity ofθ, we have
δκ(
[
α∈κ
θ(p00α))∈X.
But by the definition of p00 we haveX ⊆δP−(κ)(p00), thereforeH ◦ dID,ID◦Kewould not be a realizer of Cκ.
Now we want to prove that CκI 6≤s,w ID. Assume CIκ ≤s,w ID. Then there are H and K continuous
functions such thatH◦ dID,ID◦Keis a realizer of CIκ. As beforeH◦ dID,ID◦Keis continuous, therefore there is a Wadge strategyθ:⊆κ<κ→κ<κ such that
H◦ dID,ID◦Ke(p) = [
α∈κ
Now define the following sequence
p=Jι(wα)
_
ι(w0α)Kα∈κ,
wherewαandwα0 are such that
I0(wα) = ( 1 α, 1 2 − 1 α) and I 0(w0 α) = ( 1 3 + 1 α,1− 1 α).
Note thatδ[0,1](p) = [12,23]∪ {0,1}. Then we have:
δRκ(H◦ dID,ID◦Ke(p))∈[1 2,
2
3]∪ {0,1}. Without loss of generality we can assume
δRκ(H◦ dID,ID◦Ke(p))∈[1 2,
2 3] the other cases can be proved similarly. Letαbe such that
δRκ([θ(pα)])⊆[1 2, 2 3]. Now, define p0=Jι(wα)Kα∈κ,
where wα are such that I(wα) = (α1,1] (note that they are open intervals in [0,1]). Let p00 = (pα)_p0.
Then trivially δ[0,1](p00) ={0}. But by monotonicity of θ and by the fact thatpα=p00α, it follows that
δRκ([θ(p00α)])∈[12,23] and then
H◦ dID,ID◦Ke(p00)∈[1 2,
2 3]. But this means thatH◦ dID,ID◦Keis not a realizer of Cκ
I.
One important tool for studying the Weihrauch degrees of choice principles are the boundedness principles. In particular, reductions with boundedness principles usually result to be easier to construct and will then simplify our proofs.
Definition 4.5.5(Boundedness Principles). We define the following boundedness principles: • Bκ
I: given two Cauchy sequences,(qα)α∈κ and(qα0)α∈κ in Qκ such that
sup α∈κ qα≤ inf α∈κq 0 α,
there is x∈Rκ such that
supqα≤x≤infqα0. Formally we defineBκ I as follows: BκI :⊆R<κ×R > κ ⇒Rκ,(x, y)7→[x, y] dom(BI) ={(x, y)|x≤y}.
• Bκ: given a Cauchy sequence(qα)α∈κ of κ-rationals such that
sup
α∈κ
qα≤rfor somer∈Rκ.
Intuitively,Bκ chooses a κ-real which is greater than or equal to supα∈κqα. Formally:
Now we will focus on proving that as in the classical case we have Bκ≡wCκ and CIκ≡s,wBκI.
The formal proofs of these facts can be slightly convoluted. For this reason, before we prove these results we will give an informal intuition of the reason why they work.
First we will prove that Bκ ≡wCκ. While it is not complicated to prove that Bκ≤wCκ, the proof for
the other direction is less intuitive.
We are given an enumeration of a setA⊂κand we want to find an element inκ\A by using a realizer of Bκ. To find this element, we read the enumeration of Aand we build two sequences of rational numbers
hp andfp. The sequencehp will keep track of the smallest ordinal number not in the portion ofAwe have
read so far. The sequencefp, instead, will be a strictly Cauchy sequence whose limit is the position in which
hpbecomes constant. Intuitively hp is defined as follows:
Start by settinghp(0). Then look ata0, namely the first element of the enumeration ofA. Ifa0>0 then
hp(1) =a0 otherwisehp(1) =hp(0). In general for aα we havehp(α) =aα ifaα > hp(β) for everyβ < α,
andhp(α) =Sβ∈αhp(β) otherwise.
By using this sequence we can define fp. Intuitively fp will guess the smallest position on which hp
stabilizes.
We start definingfp(0) =−12, guessing that thehpis the constant 0. Now we checkhp(1), if it is 0, then
our guess is sill valid and we set fp(1) =−13. Otherwise we guess thathp stabilizes at position 1 and set
fp(1) = 13. In general, we check hp(α), if this does not contradict our guess g we keep defining a sequence
converging atgby settingfp(α) =g−α+11 , otherwise we change our guess toαand we setfp(α) =α−α+11 .
Note that since κ\Ais nonempty, then fp is Cauchy. Now we can feed a realizer of Bκ with hp and it
will find an upper bound r of fp. The κ-real r will be given to us as a fast converging Cauchy sequence.
Therefore we can easily find an ordinalαm which is bigger thanr. Now we can go through the enumeration
ofAagain and find the least upper bound of the firstαm elements ofA. This number will be by definition
inκ\Aas desired.
Proposition 4.5.6. Bκ≡wCκ.
Proof. First we will prove Cκ ≤w Bκ. Let pbe an enumeration of the set A. We want to find an element
not inA. We define the following sequence:
hp(0) = 0,
hp(α) = minκ\ {p(β)|β ≤α}.
Intuitivelyhp(α) is the minimum ordinal which is not inAaccording to the information inp(α+ 1). Now
by usinghp we can define the following sequence:
sp(0) = 0, sp(α+ 1) = ( sp(α) Ifhp(α) =hp(α+ 1), α+ 1 otherwise, sp(λ) = [ α∈λ sp(α).
The intuition is thatspkeeps track of the changes inhp. SinceAis non empty,spis increasing and eventually
constant, hence Cauchy. In particularsp is a Cauchy sequence whose limit`is an index ofpsuch thathp(`)
is the smallest ordinal not inA. We want to extract formsp a strictly increasing subsequence. We define:
fp(α) =sp(α)−
1
α+ 1.
The Cauchy sequencefpis trivially increasing and has limit`. Therefore we can define Kas follows:
K(p) =Jι(νQ−1
Note that, for everyα, we use only a small portion ofpin order to definefp(α), thereforeKis continuous.
Note that Bκ(K(p)) is aκ-real number greater than every element inA. It is not hard to see that, since
Bκ(K(p)) is fast convergent, by considering an initial portion of Bκ(K(p)) of length less thanκwe can find
an ordinalαm which is not in A(note that is enough to pickαm such thatαm>Bκ(K(p))). Then we can
define the functionH as follows:
H(dp,Bκ(K(p))e) = minκ\ {p(α)|α < αm}_0,
where0is the constant 0 sequence inκκ. Now, note thatH is continuous andδ
κ(H(dp,Bκ(K(p))e))∈κ\A
as desired.
Now we want to show that Bκ ≤s,w Cκ. Let p be the code of a strictly increasing Cauchy sequence
(qα)α∈κof κ-rationals. We want to find aκ-real which is greater than or equal to the least upper bound of
the sequence. DefineK as follows:
K(p)(dα, βe) =
β Ifpαcodes the sequence (qγ)γ∈γ0, and there existsγ≤γ0 such thatβ ≤qγ,
γ0 otherwise,
where γ0 is the smallest ordinal such that γ0 ≤q0. Since for defining an initial segment ofK(p) we only
need a small portion ofp, we have thatK is continuous. Now define H(α) =α_0. We claim that K(p)
is a code for the set of all the ordinal numbers smaller than or equal to ` = limα∈κqα. Indeed, ifβ ≤ `
there isqαsuch thatβ≤qα. Takeγbe such thatpγcodes the sequence (q0γ)γ0∈γ00 withα≤γ00. Therefore
K(p)(dγ, βe) =β. On the other hand, if K(p)(dα, βe) =γ, eitherγ=γ0≤q0 or there is an elementqγ0 of the sequence coded bypsuch thatγ≤qγ0.
Now by using Cκ, we obtain an ordinal which is greater than or equal to`. Then we can define a continuous
functionH(p) =p0, wherep0 is any code forp(0) inRκ. Therefore we haveδRκ(H◦Cκ◦K(p))∈[`,+∞) as
desired.
We will now prove that Cκ
I ≡s,wBIκ. Let us illustrate the idea behind the proof of C
κ
I ≤s,wBIκ. We are
given a closed subinterval [r, r0] of [0,1] as the listing of all the open intervals with κ-rational end points which have empty intersection with it. We have to define two sequences (qα)α∈κand (qα0)α∈κofκ-rationals,
respectively strictly increasing and strictly decreasing such that every elementxin between supα∈κqα and
infα∈κq0αis in [r, r0]. Let us consider the construction of (qα)α∈κ, a similar reasoning works for (q0α)α∈κ. The
idea is that of building a Cauchy sequence forr. We can start by setting q0= 0. Now we start reading the
description of [r, r0] until we find enough open intervals to cover [0, q] with qany κ-rational. Then we are sure thatq < rand we can setq1=q. In general to defineqαwe read an initial portion of the code of [r, r0]
long enough to cover [0, q] with qstrictly bigger than all theqβ forβ < αand we set qα=q. It is not hard
to see that this is a strictly increasing Cauchy sequence converging toras we wanted.
Proposition 4.5.7. Cκ
I ≡s,wBIκ.
Proof. First we prove Cκ
I ≤s,w BIκ. Let p code the closed interval [r, r0] ⊂ [0,1]. In particular assume
that p is the listing of open intervals (Jα)α∈κ such that Sα∈κJα = Rκ\[r, r0]. We want to define two
Cauchy sequences (qα)α∈κ and (q0α)α∈κ in Rκ respectively strictly increasing and strictly decreasing, such
that supα∈κqα≤infα∈κqα0. Let (q00α)α∈κbe a listing ofQκ. We define
q0= 0,q00= 1, qα=qm00where m= min{β|[0, q00β]⊆ [ γ∈αm Jγ∧qβ00>{qγ |γ < α}}, q0α=qm00where m= min{β|[qβ00,1]⊆ [ γ∈α0 m Jγ∧qβ00<{qγ0 |γ < α}},
and {β |[q00β,1]⊆ [ γ∈α0 m Jγ∧qβ00<{q0γ|γ < α}} 6=∅.
Note that αm and α0m exist by definition of (Jα)α∈κ and by the fact that the κ-rationals are dense in
Rκ. Hence the sequences are well defined. Moreover, by definition, (qα)α∈κ and (q0α)α∈κ are Cauchy and
respectively strictly increasing and decreasing. We claim supα∈κqα=rand infα∈κqα0 =r0. Since every qα
is in someJγ, it follows that r is an upper bound of (qα)α<κ. Moreover let r00 < r be such that r00 is an
upper bound of (qα)α<κ. Then there is a rational r00 < q00γ0 < r, and q00γ0 ∈ Jγ for some γ < κ. Then for
α > γ there is qα = qγ0000 with γ00 > γ0. But since q00γ0 > r00 ≤(qα)α∈κ and [0, qγ000] ⊂Sβ<α
mJβ, it follows
that qα = q00γ00 implies γ00 ≤ γ0 which contradicts our assumption. A similar proof shows infα∈κq0α =r0.
Then the functionK:p7→ dp0, p00ewherep0 andp00 are respectivelyJι(ν−1
Qκ(qα))Kα∈κ andJι(ν
−1
Qκ(q
0
α))Kα∈κ is
continuous. Moreover, takingbκI any realizer of BκI we haveδRκ(b
κ
IK(p))∈[r, r0] as desired.
Now we want to prove BIκ≤s,wCIκ. First of all note that the functionf : (0,1)→Rκ defined as
f(x) = 2x−1
x−x2
is a homeomorphism between (0,1) andRκ. Indeed, sincef is definable in the language of real closed fields
and it is a strictly increasing bijection over Rit is a strictly increasing bijection over Rκ. Moreover, since
(r, r0)∈Rκ, we havef((r, r0)) = (2rr−−r12,
2r0−1
r0−r02), it follows thatf is open inRκ.
Now let pbe the code of the two Cauchy sequences (qα)α∈κ and (q0α)α∈κ, respectively strictly increasing
and strictly decreasing. Define the following sequence:
p0 =dJI0−1([0, f(qα))Kα∈κ,JI
0−1((f(q0
α),1])Kα∈κe.
Since to define a small prefix ofp0 only a small prefix ofpis needed, the functionK:p7→p0 is continuous. Now by the fact that f−1 is continuous in Rκ it has a continuous realizer F−1. Define H = F−1. Then,
givencκ
I a realizer of C
κ
I, we have that c
κ
IK(p) is the code for a real number in [f(supα∈κqα), f(infα∈κq0α)]
and thenδRκ(Hcκ
IK(p))∈[supα∈κqα,infα∈κqα0] as desired.
Now we define the following set:
U1= [
α<κ
{p∈κκ|p(α)6= 0}.
We end this section by showing that CU1 is strictly more complex than Cκ. Proposition 4.5.8. Cκ≤s,wCU1.
Proof. DefineK as follows:
K(p)(dα, βe) =
(
1 if∃γ < α. p(γ) =β,
0 otherwise, thereforeK is continuous. Moreover letH be defined as follows:
H(p) =β_0iffβ < κis the least such thatp(β) = 0,
where0is the constant 0 function inκκ. Note thatH is trivially continuous in its domain. Then for every
realizercU1 of CU1 andp∈dom(CU1) such thatpcodes the complement ofA⊂κκ, we have:
(H◦cU1◦K)(p) =β⇔cU1K(p)(β) = 0
⇔ ∀α < κ. K(p)(dα, βe) = 0
⇔ ∀γ < κ. p(γ)6=β
⇔β /∈A,